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體驗譜分解

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This work by Jephian Lin is licensed under a Creative Commons Attribution 4.0 International License.

\(\newcommand{\trans}{^\top} \newcommand{\adj}{^{\rm adj}} \newcommand{\cof}{^{\rm cof}} \newcommand{\inp}[2]{\left\langle#1,#2\right\rangle} \newcommand{\dunion}{\mathbin{\dot\cup}} \newcommand{\bzero}{\mathbf{0}} \newcommand{\bone}{\mathbf{1}} \newcommand{\ba}{\mathbf{a}} \newcommand{\bb}{\mathbf{b}} \newcommand{\bc}{\mathbf{c}} \newcommand{\bd}{\mathbf{d}} \newcommand{\be}{\mathbf{e}} \newcommand{\bh}{\mathbf{h}} \newcommand{\bp}{\mathbf{p}} \newcommand{\bq}{\mathbf{q}} \newcommand{\br}{\mathbf{r}} \newcommand{\bx}{\mathbf{x}} \newcommand{\by}{\mathbf{y}} \newcommand{\bz}{\mathbf{z}} \newcommand{\bu}{\mathbf{u}} \newcommand{\bv}{\mathbf{v}} \newcommand{\bw}{\mathbf{w}} \newcommand{\tr}{\operatorname{tr}} \newcommand{\nul}{\operatorname{null}} \newcommand{\rank}{\operatorname{rank}} %\newcommand{\ker}{\operatorname{ker}} \newcommand{\range}{\operatorname{range}} \newcommand{\Col}{\operatorname{Col}} \newcommand{\Row}{\operatorname{Row}} \newcommand{\spec}{\operatorname{spec}} \newcommand{\vspan}{\operatorname{span}} \newcommand{\Vol}{\operatorname{Vol}} \newcommand{\sgn}{\operatorname{sgn}} \newcommand{\idmap}{\operatorname{id}} \newcommand{\am}{\operatorname{am}} \newcommand{\gm}{\operatorname{gm}} \newcommand{\mult}{\operatorname{mult}} \newcommand{\iner}{\operatorname{iner}}\)

from lingeo import random_int_list

Main idea

Let \(A\) be an \(n\times n\) matrix and
\(f_A : \mathbb{R}^n\rightarrow\mathbb{R}^n\) the corresponding linear function defined by \(f({\bf v}) = A{\bf v}\). Let \(\mathcal{E}_n\) be the standard basis of \(\mathbb{R}^n\).
Then \([f_A] = [f_A]_{\mathcal{E}_n}^{\mathcal{E}_n} = A\).

Let \(\beta\) be another basis of \(\mathcal{R}^n\) and \(Q = [\operatorname{id}]_\beta^{\mathcal{E}_n}\).
Then \([f_A]_\beta^\beta = Q^{-1}AQ\).

Spectral theorem (vector version)

Let \(A\) be an \(n\times n\) symmetric matrix.
Then there is an orthonormal basis \(\beta\) of \(\mathbb{R}^n\) such that \([f_A]_\beta^\beta = D\) is a diagonal matrix.
That is, there is an orthogonal matrix \(Q\) such that \(Q^\top AQ = D\) is a diagonal matrix.

Let \(\beta = \{ {\bf v}_1, \ldots, {\bf v}_n \}\) be the basis in the spectral theorem.
Then \(Q\) is the matrix whose columns are vectors in \(\beta\).
Since \(\beta\) is orthonormal, \(Q\) is orthogonal and \(Q^{-1} = Q^\top\).

Suppose the \(D\) matrix in the spectral theorem is \[\begin{bmatrix} \lambda_1 & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & \lambda_n \\ \end{bmatrix}. \]
By examining \(AQ = QD\), we have
\[AQ = A\begin{bmatrix} | & ~ & | \\ {\bf v}_1 & \cdots & {\bf v}_n \\ | & ~ & | \end{bmatrix} = QD = \begin{bmatrix} | & ~ & | \\ {\bf v}_1 & \cdots & {\bf v}_n \\ | & ~ & | \end{bmatrix} \begin{bmatrix} \lambda_1 & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & \lambda_n \end{bmatrix} = \begin{bmatrix} | & ~ & | \\ \lambda_1{\bf v}_1 & \cdots & \lambda_n{\bf v}_n \\ | & ~ & | \end{bmatrix}. \] Therefore, \(A{\bf v}_i = \lambda_i {\bf v}_i\) for \(i = 1,\ldots, n\).

If a nonzero vector \({\bf v}\) satisfies \(A{\bf v} = \lambda{\bf v}\) for some scalar \(\lambda\), then \({\bf v}\) is called an eigenvector of \(A\) and \(\lambda\) is called an eigenvalue of \(A\).

On the other hand, we may write \(A = QDQ^\top\).
Thus,
\[ A = QDQ^\top = \begin{bmatrix} | & ~ & | \\ {\bf v}_1 & \cdots & {\bf v}_n \\ | & ~ & | \end{bmatrix} \begin{bmatrix} \lambda_1 & ~ & ~ \\ ~ & \ddots & ~ \\ ~ & ~ & \lambda_n \\ \end{bmatrix} \begin{bmatrix} - & {\bf v}_1^\top & - \\ ~ & \vdots & ~\\ - & {\bf v}_n^\top & - \end{bmatrix} = \sum_{i = 1}^n \lambda_i {\bf v}_i{\bf v}_i^\top. \]

Suppose \(\{\lambda_1,\ldots,\lambda_n\}\) only has \(q\) distinct values \(\{\mu_1,\ldots, \mu_q\}\).
For each \(j = 1,\ldots, q\), we may let \(\displaystyle P_j = \sum_{\lambda_i = \mu_j} {\bf v}_i{\bf v}_i^\top\).
Thus, we have the following.

Spectral theorem (projection version)

Let \(A\) be an \(n\times n\) symmetric matrix.
Then there are \(q\) distinct values \(\mu_1,\ldots, \mu_q\) and \(q\) projection matrices \(P_1,\ldots, P_q\) such that

  • \(A = \sum_{j=1}^q \mu_j P_j\),
  • \(P_iP_j = P_jP_i\) for any \(i\) and \(j\), and
  • \(\sum_{j=1}^q P_j = I_n\).

Side stories

  • quadratic form
  • differential equation
  • diagonalization for general matrices

Experiments

Exercise 1

執行以下程式碼。

### code
set_random_seed(0)
print_ans = False
n = 3
Q = matrix([
    [1 / sqrt(3), 1 / sqrt(2), 1 / sqrt(6)],
    [1 / sqrt(3), -1 / sqrt(2), 1 / sqrt(6)],
    [1 / sqrt(3), 0, -2 / sqrt(6)]
])
v = random_int_list(n)
D = diagonal_matrix(v)
A = Q * D * Q.transpose()

cs = random_int_list(n)

print("A =")
show(A)
for i in range(n):
    print("v%s ="%(i+1), Q.column(i))
print("b = " + " + ".join("%s v%s"%(cs[i], i+1) for i in range(n))) 

if print_ans:
    for i in range(n):
        print("A v%s = %s v%s"%(i+1, v[i], i+1))
    print("A b = " + " + ".join("%s v%s"%(cs[i]*v[i], i+1) for i in range(n))) 
    print("Q =")
    show(Q)
    print("D =")
    show(D)

set_random_seed(0),得
\[ A = \begin{bmatrix} 1 & -2 & -3\\ -2 & 1 & -3\\ -3 & -3 & 2 \end{bmatrix}, \bv_1 = \begin{bmatrix} \frac{1}{3\sqrt{3}}\\ \frac{1}{3\sqrt{3}}\\ \frac{1}{3\sqrt{3}} \end{bmatrix}, \bv_2 = \begin{bmatrix} \frac{1}{2\sqrt{2}}\\ -\frac{1}{2\sqrt{2}}\\ 0 \end{bmatrix}, \bv_3 = \begin{bmatrix} \frac{1}{6\sqrt{6}}\\ \frac{1}{6\sqrt{6}}\\ -\frac{1}{3\sqrt{6}} \end{bmatrix}, \bb = -5\bv_1 -5\bv_2 + 0\bv_3 \]

Exercise 1(a)

驗證 \({\bf v}_1, \ldots, {\bf v}_3\)\(A\) 的特徵向量﹐並找出相對應的特徵值。
\(Ans:\)
直接驗證
\[ \begin{aligned} &A\bv_1 = \begin{bmatrix} -\frac{4\sqrt{3}}{9}\\ -\frac{4\sqrt{3}}{9}\\ -\frac{4\sqrt{3}}{9} \end{bmatrix}=\lambda_1\bv_1,\ \lambda_1 = -4,\\ &A\bv_2 = \begin{bmatrix} \frac{3\sqrt{2}}{4}\\ -\frac{3\sqrt{2}}{4}\\ 0 \end{bmatrix}=\lambda_2\bv_2,\ \lambda_2 = 3,\\ &A\bv_3= \begin{bmatrix} \frac{5\sqrt{6}}{36}\\ \frac{5\sqrt{6}}{36}\\ -\frac{5\sqrt{6}}{18} \end{bmatrix}=\lambda_3\bv_3,\ \lambda_3 = 5. \end{aligned} \]

Exercise 1(b)

\(A{\bf b}\) 寫成 \(\{{\bf v}_1, \ldots, {\bf v}_3\}\)的線性組合。
\(Ans:\)
利用 1(a)
\[ \begin{aligned} A\bb &= A(-5\bv_1-5\bv_2+0\bv_3)\\ &= -5A\bv_1 - 5A\bv_2\\ &= 20\bv_1 - 15\bv_2. \end{aligned} \]

Exercise 1©

找出一個垂直矩陣 \(Q\) 和一個對角矩陣 \(D\) 使得 \(D = Q^\top AQ\)

  • Orthonormal > orthonormal
  • 標點

\(Ans:\)
\(Q = \begin{bmatrix} | & | & |\\ \bv_1 & \bv_2 & \bv_3\\ | & | & | \end{bmatrix}\) 因為\(Q\)是垂直矩陣,所以 \(\beta\) 是 orthonormal basis,所以分別將 \(\bv_i\) 單位化,就可以得到 \(\bu_i\)
做單位化得
\[ Q = \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}}\\ \end{bmatrix} .\] 驗證 \(D =\begin{bmatrix} \lambda_1 & 0 & 0\\ 0 & \lambda_2 & 0\\ 0 & 0 & \lambda_3 \end{bmatrix}= Q\trans AQ\)
\[ \begin{aligned} Q\trans AQ &= \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0\\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{6}} & -\frac{2}{\sqrt{6}}\\ \end{bmatrix} \begin{bmatrix} 1 & -2 & -3\\ -2 & 1 & -3\\ -3 & -3 & 2 \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \end{bmatrix}\\ &=\begin{bmatrix} -\frac{4\sqrt{3}}{3} & -\frac{4\sqrt{3}}{3} & -\frac{4\sqrt{3}}{3}\\ -\frac{3\sqrt{2}}{2} & \frac{3\sqrt{2}}{2} & 0\\ -\frac{5\sqrt{6}}{6} & -\frac{5\sqrt{6}}{6} & \frac{5\sqrt{6}}{3} \end{bmatrix} \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}}\\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}}\\ \end{bmatrix}\\ &=\begin{bmatrix} -4 & 0 & 0\\ 0 & 3 & 0\\ 0 & 0 & 5 \end{bmatrix}. \end{aligned} \]

Exercises

Exercise 2

\(A\) 為一 \(3\times 3\) 矩陣而
\(\beta = \{ {\bf v}_1,\ldots,{\bf v}_3 \}\)\(\mathbb{R}^3\) 的一組基底。
已知
\[[f_A]_\beta^\beta = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \\ \end{bmatrix}. \]
\(A{\bf v}_1\)\(A{\bf v}_2\)\(A{\bf v}_3\)、及 \(A({\bf v}_1 + {\bf v}_2 + {\bf v}_3)\) 分別寫成 \(\beta\) 的線性組合。

  • 前兩行看起來沒用到?
  • \([A]_\beta^\beta\) > \([f_A]_\beta^\beta\)
  • 矩陣右下角的 \(\beta\) 拿掉,這門課沒有用這個符號

\(Ans:\)
\([A {\bf v}_1]_\beta = [f_A]_\beta ^\beta [{\bf v}_1]_\beta = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ 0 \end{bmatrix} = 3 {\bf v}_1.\)
\([A {\bf v}_2]_\beta = [f_A]_\beta ^\beta [{\bf v}_2]_\beta = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 4 \\ 0 \end{bmatrix} = 4 {\bf v}_2.\)
\([A {\bf v}_3]_\beta = [f_A]_\beta ^\beta [{\bf v}_3]_\beta = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 5 \end{bmatrix} = 5 {\bf v}_3.\)
\(A({\bf v}_1 + {\bf v}_2 + {\bf v}_3) = A {\bf v}_1 + A {\bf v}_2 + A {\bf v}_3 = 3 {\bf v}_1 + 4 {\bf v}_2 + 5 {\bf v}_3.\)

Exercise 3

\[A = \begin{bmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{bmatrix} \] \(\beta = \{ {\bf v}_1, \ldots, {\bf v}_3 \}\)
\[\begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \] 的行向量集合。

Exercise 3(a)

寫出 \([f_A]_\beta^\beta\) 並說明 \(f_A\) 的作用。

\(Ans:\)

\[[f_A]_\beta^\beta = \begin{bmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \\ \end{bmatrix}. \]

\(f_A\) 的作用為:
將基底 \(\beta\) 分解成 \(c_1{\bf u}_1 + c_2{\bf u}_2 + c_3{\bf u}_3\) 的形式,
再根據運算做伸縮,
最後再將全部一起合併。

以這題為例:
將基底 \(\beta\) 分解成 \(c_1{\bf u}_1 + c_2{\bf u}_2 + c_3{\bf u}_3\)
經過運算伸縮及合併後可得\(\beta = 2c_1{\bf u}_1-c_2{\bf u}_2-c_3{\bf u}_3\)

Exercise 3(b)

找出一個垂直矩陣 \(Q\) 和一個對角矩陣 \(D\) 使得 \(D = Q^\top AQ\)

  • \(D\) 有打錯

\(Ans:\)

\(Q\) 為從標準基底到 \(\beta\) 的基底變換矩陣,
則可得: \[Q= \begin{bmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \\ \end{bmatrix} \]

\[D =\begin{bmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix} \]

Exercise 3©

\(P_1\) 為投影到 \(\operatorname{span}(\{{\bf v}_1\})\) 的投影矩陣、
\(P_2\) 為投影到 \(\operatorname{span}(\{{\bf v}_2, {\bf v}_3\})\) 的投影矩陣。
說明 \(P_1 = {\bf v}_1{\bf v}_1^\top\)\(P_2 = {\bf v}_2{\bf v}_2^\top + {\bf v}_3{\bf v}_3^\top\)

  • 中英數間空格
  • 同理,> 另一部份,因為 \(\bv_2\)\(\bv_3\) 為互相垂直的單位向量,所以一個向量對 \(\vspan\{\bv_2,\bv_3\}\) 的投影,等同於該向量投影到 \(\vspan\{\bv_2\}\) 和投影到 \(\vspan\{\bv_3\}\) 的結果相加,所以

\(Ans:\)

若有一個向量 \(u\) 進來和 \({\bf v}_1^\top\) 作內積,
內積完再乘以 \({\bf v}_1\)
所以合在一起就是內積乘以 \({\bf v}_1\),等於投影矩陣,
所以 \(P_1 ={\bf v}_1{\bf v}_1^\top\)即為 \(\operatorname{span}(\{{\bf v}_1\})\)的投影矩陣,
另一部份,因為 \(\bv_2\)\(\bv_3\) 為互相垂直的單位向量,
所以一個向量對 \(\vspan\{\bv_2,\bv_3\}\) 的投影,
等同於該向量投影到 \(\vspan\{\bv_2\}\) 和投影到 \(\vspan\{\bv_3\}\) 的結果相加,
所以 \(P_2 = {\bf v}_2{\bf v}_2^\top + {\bf v}_3{\bf v}_3^\top\) 即為 \(\operatorname{span}(\{{\bf v}_2, {\bf v}_3\})\) 的投影矩陣。

Exercise 3(d)

\(A\) 寫成一些投影矩陣的線性組合﹐並再次說明 \(f_A\) 的作用﹐看看是否和第一小題一致。

  • \({\bf v}_2{\bf v}_2^\top + {\bf v}_3{\bf v}_3^\top\)\(\operatorname{span}(\{{\bf v}_2\})\)的投影矩陣。< 這句有錯

\(Ans:\)

\(A =2{\bf v}_1{\bf v}_1^\top - {\bf v}_2{\bf v}_2^\top - {\bf v}_3{\bf v}_3^\top\)
其中 \({\bf v}_1{\bf v}_1^\top\)\(\operatorname{span}(\{{\bf v}_1\})\)的投影矩陣,
\({\bf v}_2{\bf v}_2^\top + {\bf v}_3{\bf v}_3^\top\)\(\operatorname{span}(\{{\bf v}_2, {\bf v}_3\})\)的投影矩陣。
\(f_A\) 的作用仍是把向量先分解,再伸縮,最後合併,
和第一小題的作用一致。

Exercise 4


\[A = \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ \end{bmatrix}. \]

Exercise 4(a)

說明要找一個非零向量 \({\bf v}\) 使得 \(A{\bf v} = \lambda{\bf v}\)
等同於在 \((A - \lambda I){\bf v} = {\bf 0}\) 中找非零解。

  • 因為 \((A - \lambda I){\bf v} = {\bf 0}\) 為一個方程式的表示方法, > 我們可以將 \(\lambda\bv\) 寫成 \(\lambda I\bv\),則 \(A\bv = \lambda I\bv\)\((A - \lambda I) \bv = \bzero\) 等價。

Ans:
我們可以將 \(\lambda\bv\) 寫成 \(\lambda I\bv\),則 \(A\bv = \lambda I\bv\)\((A - \lambda I) \bv = \bzero\) 等價。
又要求 \({\bf v}\) 為非零向量,所以就等同於在 \((A - \lambda I){\bf v} = {\bf 0}\) 中找非零解。

Exercise 4(b)

方程式 \((A - \lambda I){\bf v} = {\bf 0}\) 有非零解只會發生在 \(\det(A - \lambda I) = 0\) 的時候。
利用這個性質找出所有可能的 \(\lambda\)

Ans:
\(\det(A - \lambda I) = 0\) 可算得 \(\lambda(\lambda-5)=0\)
求得 \(\lambda=0\)\(5\)

Exercise 4©

對每一個 \(\lambda\) 解出相對應的 \({\bf v}\)
向量 \({\bf v}\) 的選擇可能很多﹐把它的長度縮為 \(1\)

  • 向量粗體
  • 第二個 \(A\) 的等式有錯

Ans:
\(\lambda=0:\)
\(A-\lambda I = \begin{bmatrix} 1 & 2 \\ 2 & 4 \\ \end{bmatrix}\)\({\bf v}=(2,-1)\)
\({\bf v}'=(2/{\sqrt{5}},-1/{\sqrt{5}})\)

\(\lambda=5:\)
\(A- \lambda I = \begin{bmatrix} -4 & 2 \\ 2 & -1 \\ \end{bmatrix}\)\({\bf v}=(1,2)\)
\({\bf v}'=(1/{\sqrt{5}},2/{\sqrt{5}})\) 以上兩解為所求。

Exercise 4(d)

找出一個垂直矩陣 \(Q\) 和一個對角矩陣 \(D\) 使得 \(D = Q^\top AQ\)

  • 向量粗體
  • \(Q\) < 這裡 \(Q\) 是什麼
  • 邏輯不完整:取了這個 \(Q\) 和這個 \(D\) 之後為什麼 \(D = Q\trans AQ\)

Ans:
由4©,我們取 \(\beta=\{ v_1,v_2 \}\),其中 \({\bf v}_1=(\frac{2}{\sqrt{5}},\frac{-1}{\sqrt{5}}),{\bf v}_2=(\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}})\)
則垂直矩陣 \(Q\) 為從標準基底到 \(\beta\) 的基底變換矩陣,
也就是
\[Q= \begin{bmatrix} \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} \\ \frac{-1}{\sqrt{5}} & \frac{2}{\sqrt{5}} \\ \end{bmatrix} \]
經過直接計算可得 \[D = Q^\top AQ =\begin{bmatrix} 0 & 0 \\ 0 & 5 \\ \end{bmatrix}. \]

Exercise 5


\[A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ \end{bmatrix}, Q = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \\ \end{bmatrix}, D = \begin{bmatrix} 3 & 0 \\ 0 & -1 \\ \end{bmatrix}. \]

Exercise 5(a)

驗證 \(Q^\top AQ = D\)

Ans:
\[Q^\top AQ = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & - 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 2 & 1 \\ \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & - 1 \\ \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 1 & 1 \\ 1 & - 1 \\ \end{bmatrix} \begin{bmatrix} 3 & -1 \\ 3 & 1 \\ \end{bmatrix} = \frac{1}{2} \begin{bmatrix} 6 & 0 \\ 0 & -2 \\ \end{bmatrix} = \begin{bmatrix} 3 & 0 \\ 0 & -1 \\ \end{bmatrix} = D. \]

Exercise 5(a)

\(p(x,y) = x^2 + 4xy + y^2\)
找一些係數 \(a,b,c,d\) 並令
\(\hat{x} = a x + b y\)
\(\hat{y} = c x + d y\)
使得 \(p(x,y) = 3\hat{x}^2 - \hat{y}^2\)
藉此說明 \(p(x,y) = 1\) 的圖形是雙曲線。

這題想法不錯,不過不清楚。
我直接改寫。

Ans:
設一線性函數為
\[ f\left(\begin{bmatrix} x \\ y \end{bmatrix}\right) = \begin{bmatrix} x + 2y \\ 2x + y \end{bmatrix}. \] \[ p(x,y) = x(x+2y) + y(2x+y), \] 恰好為 \((x,y)\)\(f(x,y)\) 的內積。

已知正交的標準基底變換不影響線性函數與內積,
若能找一組正交標準基底 \(\beta\) 使得
\[ \inp{[(x,y)]_\beta}{[f(x,y)]_\beta} = 3\hat{x}^2 - \hat{y}^2, \] 其中 \([(x,y)]_\beta = (\hat{x}, \hat{y})\)
\[ p(x,y) = \inp{(x,y)}{f(x,y)} = \inp{[(x,y)]_\beta}{[f(x,y)]_\beta} = 3\hat{x}^2 + \hat{y}^2. \]

如果可以取到
\[ [f]_\beta^\beta = \begin{bmatrix} 3 & 0 \\ 0 & -1 \end{bmatrix} \] 則可以得到這個效果。

因此根據 5(a),我們可以取 \(\beta\)\(Q\) 矩陣的行向量集合。
如此一來就有
\[ \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}. \] 也就是說 \(a = \frac{1}{\sqrt{2}} , b = \frac{1}{\sqrt{2}} , c = \frac{1}{\sqrt{2}} , d = - \frac{1}{\sqrt{2}}\)

Exercise 5(b)

\(x(t), y(t)\) 為以 \(t\) 為變數的函數。
\(x'(t), y'(t)\) 為其對 \(t\) 的微分。
考慮微分方程
\(x' = x + 2y\)
\(y' = 2x + y\)
找一些係數 \(a,b,c,d\) 並令
\(\hat{x} = a x + b y\)
\(\hat{y} = c x + d y\)
使得原方程可以改寫為
\(\hat{x}' = 3\hat{x}\)
\(\hat{y}' = -\hat{y}\)
(此方程的解為
\(\hat{x} = C_1e^{3t}\)
\(\hat{y} = C_2e^{-t}\)
其中 \(C_1\)\(C_2\) 是任意常數。)
解原方程。

Ans:
由於 \(\begin{cases} x' = x + 2y \\ y' = 2x + y \end{cases}\)
可寫成 \(\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}\)
設一基底變換矩陣 \(B\)
會使 \(B \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix}\)
\(B \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix} B^ {-1} \begin{bmatrix} \hat{x} \\ \hat{y} \end{bmatrix} = \begin{bmatrix} 3 \hat{x} \\ -1 \hat{y} \end{bmatrix}\)
因此根據5(a),
\(B = \begin{bmatrix} a & b \\ c & d \end{bmatrix} = Q^ \top = \begin{bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & - \frac{1}{\sqrt{2}} \end{bmatrix}\)
也就是說, \(a = \frac{1}{\sqrt{2}} , b = \frac{1}{\sqrt{2}} , c = \frac{1}{\sqrt{2}} , d = - \frac{1}{\sqrt{2}}\)

Exercise 6

以下例題說明並非對稱矩陣才能表示成對角矩陣。
然而其所用的基底不再是垂直的。

\[A = \begin{bmatrix} 1 & 2 \\ 1 & 2 \\ \end{bmatrix}. \]

Exercise 6(a)

說明要找一個非零向量 \({\bf v}\) 使得 \(A{\bf v} = \lambda{\bf v}\)
等同於在 \((A - \lambda I){\bf v} = {\bf 0}\) 中找非零解。

  • 向量粗體
  • 也就是那句刪掉

\(Ans:\)
\(A{\bf v}\) = \(\lambda{\bf v}\) 可寫成 \((A-\lambda I){\bf v}\) = \(0\) ,
由於 \({\bf v}\) 為非零向量 , 所以也是非零解 。

Exercise 6(b)

方程式 \((A - \lambda I){\bf v} = {\bf 0}\) 有非零解只會發生在 \(\det(A - \lambda I) = 0\) 的時候。
利用這個性質找出所有可能的 \(\lambda\)

  • 句子要完整,不要只有數學式
  • \(\lambda = 0 或是 3\) > \(\lambda\)\(0\) 或是 \(3\)

\(Ans:\)
直接計算可得
\(A - \lambda I\) = \(\begin{bmatrix}1 & 2 \\ 1 & 2 \end{bmatrix} -\begin{bmatrix} \lambda & 0 \\ 0 & \lambda\end{bmatrix} = \begin{bmatrix} 1 - \lambda & 2 \\ 1 & 2 - \lambda \end{bmatrix}.\)
\(\det(A - \lambda I)\)\(0\) 時,
行列式值為 \((1 - \lambda)(2 - \lambda) - 2 = 0\) ,
也就是 \(\lambda(\lambda - 3) = 0\) ,
\(\lambda\)\(0\) 或是 \(3\)

Exercise 6©

對每一個 \(\lambda\) 解出相對應的 \({\bf v}\)
向量 \({\bf v}\) 的選擇可能很多﹐把它的長度縮為 \(1\)

  • 中英數間空格

\(Ans:\)
假設 \({\bf v}\) = \[\begin{bmatrix}a \\ b \end{bmatrix}\]
\(\lambda = 0\) ,
\(\begin{bmatrix} 1 & 2 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}\)
因為 \(a\) = \(-2 b\) , 所以 \(a = \frac{-2}{\sqrt{5}} , b = \frac{1}{\sqrt{5}}\)
\(\lambda = 3\) ,
\(\begin{bmatrix} 1 & 2 \\ 1 & 2\end{bmatrix} \begin{bmatrix} a \\ b \end{bmatrix} = 3 \begin{bmatrix} a \\b \end{bmatrix}\)
因為 \(a = b\) , 所以 \(a = \frac{1}{\sqrt{2}} , b = \frac{1}{\sqrt{2}}\)

Exercise 6(d)

找出一個可逆矩陣 \(Q\) 和一個對角矩陣使得 \(D = Q^{-1} AQ\)

  • 數學進數學式
  • \(V(0)\) 是什麼,第二行第三行的數字又是哪來的。
  • 等號進數學式

這題要用上一題,目前的答案沒有解釋為什麼 \(Q^{-1}AQ = D\)

\(Ans:\)
由 6(b) 知特徵根為 \(0\)\(3\)
並由上一題的計算可得到,
\(\operatorname{ker}(A - 0 I)\) = \(\operatorname{span}\left\{\begin{bmatrix}-2 \\ 1\end{bmatrix}\right\}\)\(A\) 相對於 0 的特徵空間 ,
\(\operatorname{ker}(A - 3 I)\) = \(\operatorname{span}\left\{\begin{bmatrix}1 \\ 1\end{bmatrix}\right\}\)\(A\) 相對於 3 的特徵空間 ,
所以 \(\begin{bmatrix}-2 \\ 1\end{bmatrix}\) , \(\begin{bmatrix}1 \\ 1\end{bmatrix}\)\(A\) 的兩個線性獨立的特徵向量。

由於
\[ A\begin{bmatrix} -2 \\ 1 \end{bmatrix} = 0 \begin{bmatrix} -2 \\ 1 \end{bmatrix} \] \[ A\begin{bmatrix} 1 \\ 1 \end{bmatrix} = 3 \begin{bmatrix} 1 \\ 1 \end{bmatrix}. \] 所以當 \(\beta\) 是由這兩個向量所組成的基底時, \[ [f_A]_\beta^\beta = \begin{bmatrix} 0 & 0 \\ 0 & 3 \end{bmatrix} = [\operatorname{id}]_{\mathcal{E}_2}^\beta [f_A] [\operatorname{id}]_{\beta}^{\mathcal{E}_2}. \] 因此可以取 \(Q = [\operatorname{id}]_{\beta}^{\mathcal{E}_2}\)= \(\begin{bmatrix}-2 & 1 \\1 & 1\end{bmatrix}\) , \(Q^{-1}AQ\) = \(D\) = \(\begin{bmatrix}0 & 0\\ 0 & 3\end{bmatrix}\)

目前分數 4.5

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